Popular Mechanics - USA (2020-07)

(Antfer) #1

Deep Math


We Spent All


Day Arguing


About This


Triangle Brain


Teaser. Can


You Solve It?


N


O T H I N G H A LT S P R O D U C T I V I T Y AMONG
coworkers quite like a maddening brain
teaser. The latest to ensnare the Popu-
lar Mechanics editors and readers: How
many triangles are in this drawing?
When I posed the problem to our
team, responses ranged from 4 trian-
gles all the way to 22. Most people saw 18. One
wiseguy counted the triangles in the A’s in the
question itself, while another seemed to have an
existential crisis: “None of these lines are truly
straight, just curves—thus you cannot define any
of them as a triangle,” he said. “There are no tri-
angles in this photo. Life has no meaning.”
I could’ve listened to my colleagues’ question-
able processes all day, but instead, I reached out
to geometry experts to see if we could arrive at
a consensus. All of the mathematicians I con-
tacted found the same solution—but not all of
them figured it out in the same way.
“I would approach this like one approaches
any mathematical problem: reduce it and find
structure,” says Sylvester Eriksson-Bique, PhD,
a postdoctoral fellow with the University of Cal-
ifornia Los Angeles’s math department.
The only way to form triangles in the draw-
ing, Eriksson-Bisque says, is if the top vertex
(corner) is part of the triangle. The base of the
triangle will then have to be one of the three lev-
els below. “There are three levels, and on each
you can choose a base among six different ways.
This gives 18, or 3 times 6 triangles.”
Look at the triangle again: “It’s convenient to
generalize to the case where there are ‘n’ lines
passing through the top vertex, and ‘p’ hori-
zontal lines,” says Francis Bonahon, PhD, a
professor of mathematics at the University of
Southern California.
In our case, n = 4, and p = 3. Any triangle we
find in the drawing should have one top vertex
and two others on the same horizontal line, so
for each horizontal line, the number of triangles
is equal to the number of ways we can choose two
distinct vertices on that line out of n total points,
Bonahon says—or “n choose 2.”
That’s n(n–1)/2. And since there are p hor-
izontal lines, this gives p*n(n–1)/2 possible
triangles. In our case, that’s 3*4(4–1)/2 = 18.
Here’s a breakdown of how to find each pos-
sible triangle:

// BY A NDRE W DA NIELS //

6

Free download pdf